To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.
PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

Where,
G = Gravitational Universal Constant
d = Distance
M = Mass
Radius earth center of mass
PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,



PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

At the same time we have that centripetal acceleration is given as

Replacing



In short, and in general:
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Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides

i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by


(b)
Magnetic flux,





(c)
R = 3 ohm

magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
I mean if he flies 5g that means that's his average speed too